It is known that in the triangle ABC of RT, the angle ACB = 90 degrees, D and E are the midpoint of AB and BC respectively, and the point F is on the extension line of AC, and CF = De

It is known that in the triangle ABC of RT, the angle ACB = 90 degrees, D and E are the midpoint of AB and BC respectively, and the point F is on the extension line of AC, and CF = De

Since D and E are the midpoint of AB and BC respectively, De is the median line of the triangle ABC, de ∥ CF, and de = CF is known,
So the quadrilateral DEFC is a parallelogram,
∴CD‖EF.

As shown in the figure △ ABC is an equilateral triangle, D and F are points on AB and BC respectively, e is a point outside △ ABC, de = CF, EF = DC, the extended line of ED intersects AC with C, EC = AC One is to verify; the other is to judge the shape of △ AEF

Connect the DF,
In △ EFD and △ DFC
∵DE＝CF,EF＝DC,DF＝DF
∴△EFD≌△DFC
| EDFC is a parallelogram, eg|||bc
∴AD＝AG,∠AGD＝∠GAD
And ∵ eg = AC
≌△ DAC
∴AE＝CD＝EF
The △ AEF is an isosceles triangle

In the right angle ABCD, AB / / DC, ∠ ABC = 90 °, ab = 2CD, the correction point of diagonal AC and BD is f, and the crossing point F is EF / / AB, and the intersection of ad with E is made Verification: the quadrilateral affe is isosceles trapezoid It's right angle trapezoid ABCD

prove:
Do DG ⊥ AB through point d
Because the angle ABC is equal to 90 degrees
So CB ⊥ ab
So DG ‖ BC
Because CD ∥ ab
So the quadrilateral GBCD is a rectangle
So CD = GB
Because AB = 2CD
So AB = 2GB
So G is the midpoint of ab
That is, DG divides AB vertically
So ad = BD (the distance from the point on the vertical bisector of the line segment to both ends of the line segment is equal)
So angle DAB = angle DBA
Because EF ‖ ab
So the quadrilateral Abe is isosceles trapezoid

As shown in the figure, in the right angle trapezoid ABCD, ab ∥ DC, ∠ ABC = 90 °, ab = 2dc, diagonal AC ⊥ BD, vertical foot F, EF ∥ AB, crossing ad at point E, CF = 4cm (1) The results show that the quadrilateral AFE is isosceles trapezoid; (2) Find the length of AE

(1) It is proved that the crossing point D is DM ⊥ ab,
∵DC∥AB，∠CBA=90°，
The quadrilateral BCDM is rectangular
∴DC=MB．
∵AB=2DC，
∴AM=MB=DC．
∵DM⊥AB，
∴AD=BD．
∴∠DAB=∠DBA．
∵ EF ∫ AB, AE and BF intersect at point D, that is, AE and FB are not parallel,
The quadrilateral affe is isosceles trapezoid
（2）∵DC∥AB，
∴△DCF∽△BAF．
∴CD
AB=CF
AF=1
2．
∵CF=4cm，
∴AF=8cm．
∵AC⊥BD，∠ABC=90°，
In △ ABF and △ BCF,
∵∠ABC=∠BFC=90°，
∴∠FAB+∠ABF=90°，
∵∠FBC+∠ABF=90°，
∴∠FAB=∠FBC，
｛ Abf ∽ BCF (AA), namely BF
CF=AF
BF，
∴BF2=CF•AF．
∴BF=4
2cm．
∴AE=BF=4
2cm．

As shown in the figure, in the right angle trapezoid ABCD, ab ∥ DC, ∠ ABC = 90 °, ab = 2dc, diagonal AC ⊥ BD, vertical foot F, EF ∥ AB, crossing ad at point E, CF = 4cm (1) The results show that the quadrilateral AFE is isosceles trapezoid; (2) Find the length of AE

(1) It is proved that: the crossing point D is DM ⊥ AB, ∵ DC ∥ AB, ∵ CBA = 90 ° and ᙽ BCDM is a rectangle.

It is known that: AB is the diameter of circle O, CD is the chord, AE ⊥ CD, the perpendicular foot is e, BF ⊥ CD, and the perpendicular foot is F. verification: EC = DF if AE = 1, BF = 7, the radius of circle O is 5, and find the length of CD

Tips:
Make om ⊥ CD at point M
OM=1/2(7-1)=3
CM²=OC²-OM²=5²-3²=16
CM=4
∴CD=8

AB is the diameter of ⊙ o, CD is the chord, EC ⊥ CD, FD ⊥ CD, EC and DF intersect the diameters of AB at two points of EF respectively. It is proved that AE = BF

I haven't done geometry problem for more than 10 years. There's a way. If you don't have a better one, you can use it
Make auxiliary line: make the diameter of a circle parallel to the chord, intersect EC with G, and intersect DF with H. according to corner angle theorem, triangle OEG is equal to triangle ofH, so OE = of, because OA = ob, oa-oe = ob-of, that is, AE = BF

As shown in the figure, CD is the chord of circle O, e and F are on diameter AB, EC ⊥ CD, FD ⊥ CD verification: AE = BF (2) when the chord CD intersects with diameter AB, other conditions remain unchanged, and the conclusion is tenable As shown in the figure, CD is the chord of circle O, e and F are on the diameter AB, EC ⊥ CD, FD ⊥ CD verification: AE = BF (2) when the chord CD intersects with diameter AB, other conditions remain unchanged, is the conclusion Tenable? Try to draw a figure without proving (3) if the condition EC ⊥ CD, FD ⊥ CD is changed to AE ⊥ CD, BF ⊥ CD, AE, BF intersect CD with CD and E, f respectively, is the conclusion still tenable? Try to draw the figure and prove it

(1) if oh ⊥ CD is in H, then ch = DH, ∵ CE ⊥ CD, DF ⊥ CD, ᙽ CE ∥ oh ∥ DF,  OE / of = ch / CH = 1, and OA = ob,  AE = BF

AB is the diameter of ⊙ o, CD is the chord, EC ⊥ CD, FD ⊥ CD, respectively. The vertical feet are C and D= BF.BF Vertical CD and F, CE = BF

Pass through point o as OM, perpendicular to CD and m
According to the vertical diameter theorem, CM = DM
Because CE ⊥ CD, DF ⊥ CD
∴CE‖OM‖DF
∴OE=OF
∵O=OB
∴AE=BF

As shown in the figure, AB is the diameter of circle O, chord CD intersects AB, AE is vertical CD, BF is vertical CD, and vertical foot is e and f respectively

It is proved that if OM is perpendicular to m, then cm = DM
Connect EO and extend it to meet the straight line of BF at n
Then AE ∥ om ∥ BF
Therefore: EM / FM = EO / no = AO / Bo = 1, then EM = FM
Therefore, cm-em = dm-fm, that is, CE = DF